为什么输出不随用户的输入而改变？

Question

所以我在做我的功课，我需要做的编程挑战之一就是周期表挑战。我不认为编程会那么难，但我遇到了一个我无法解决的问题。这不是一个语法错误，我已经花了数小时在互联网上浏览了有关我是否正确使用if语句的指南，但我无法识别错误。我的代码是：

lithium = ("""Name: Lithium
Atmomic mass: 6.94
Group: Alkali Metals""")

sodium = ("""Name: Sodium
Atomic mass: 22.989769 u ± 2 × 10^-8 u
Group: Alkali Metals""")

potassium = ("""Name: Potassium
Atmoic mass: 39.0983 u ± 0.0001 u
Group: Alkali Metals""")

rubidium = ("""Name: Rubidium
Atomic mass: 85.4678 u ± 0.0003 u
Group: Alkali Metals""")

cesium = ("""Name: Ceasium
Atomic mass: 132.90545 u ± 2 × 10^-7 u
Group: Alkali Metals""")

francium = ("""Name: Francium
Atomic mass: 223 u
Group: Alkali Metals""")

################################################################################################################################################################

helium = ("""Name: Helium
Atomic Mass: 4.002602 u ± 0.000002 u
Group: Noble Gasses""")

neon = ("""Name: Neon
Atomic Mass: 20.1797 u ± 0.0006 u
Group: Noble Gasses""")

argon = ("""Name: Argon
Atomic Mass: 39.948 u ± 0.001 u
Group: Noble Gasses""")

krypton = ("""Name: Krypton
Atomic Mass: 83.798 u
Group: Noble Gasses""")

xenon = ("""Name: Xenon
Atomic Mass: 131.293 u ± 0.006 u
Group: Noble Gasses""")

radon = ("""Name: Radon
Atomic Mass: 222 u
Group: Noble Gasses""")

def noble():
    print(helium)
    print(" ")
    print(neon)
    print(" ")
    print(argon)
    print(" ")
    print(krypton)
    print(" ")
    print(xenon)
    print(" ")
    print(radon)

def alkali():
    print(lithium)
    print(" ") 
    print(sodium)
    print(" ") 
    print(potassium)
    print(" ") 
    print(rubidium)
    print(" ") 
    print(cesium)
    print(" ") 
    print(francium)
print ("""Please choose the element you are looking for from the list below
Lithium
Sodium
Potassium
Rubidium
Caesium
Francium
Helium
Neon
Argon
Krypton
Xenon
Radon

Alternatively you can type in the name of one of these groups to get the data for first 6 elements in the group

Noble Gasses
Alkali Metals""")

userchoice = str(input())


if userchoice == "Helium" or "helium":
    print(helium)
elif userchoice == "Neon" or "neon":
    print(neon)
elif userchoice == "Argon" or "argon":
    print(argon)
elif userchoice == "Krypton" or "krypton":
    print(krypton)
elif userchoice == "Xenon" or "xenon":
    print(xenon)
elif userchoice == "Radon" or "radon":
    print(radon)
elif userchoice == "Lithium" or "lithium":
    print(lithium)
elif userchoice == "Sodium" or "sodium":
    print(sodium)
elif userchoice == "Potassium" or "potassium" or "Potasium" or "potasium":
    print(potassium)
elif userchoice == "Rubidium" or "rubidium":
    print(rubidium)
elif userchoice == "Caesium" or "caesium" or "Cesium" or "cesium":
    print(cesium)
elif userchoice == "Francium" or "francium":
    print(francium)
elif userchoice == "Noble Gasses" or "noble gasses" or "noble" or "gasses":
    print(noble())
elif userchoice == "Alkali Metals" or "alkali metals" or "akali metals" or "Akali Metals":
    print(alkali())
else:
    print("Element not recognised")

现在我的问题是，不管我为userchoice变量输入什么，结果总是打印（helium）变量，我的意思是，即使我输入“neon”，它仍然会打印helium，即使helium变量只有在userchoice变量包含单词“helium”或“helium”时才应该打印。如下代码所示：

userchoice = str("Neon")


if userchoice == "Helium" or "helium":
    print(helium)
elif userchoice == "Neon" or "neon":
    print(neon)

我想知道是否有人会这么好，并帮助我使它，使程序实际工作，在这一点上只是为了澄清，无论输入，所有的程序输出是氦变量，它的内容如下：

helium = ("""Name: Helium
Atomic Mass: 4.002602 u ± 0.000002 u
Group: Noble Gasses""")

Answer 1

这个问题是由userchoice == "Helium" or "helium"这样的行引起的，当用or分隔时，python尝试读取两个完全不同的条件。在后面的例子中，条件只是helium，它总是返回true。所以把所有的行都改成userchoice == "Helium" or userchoice == "helium"。或者直接用userchoice.lower()